I’ve applied the same method as last time: Roy Morgan make ISP customer profiles available for purchase. At the bottom of each report’s synopsis, you’ll see that a sample size has been included (for example, the Internode customer profile is based on a sample of 404 customers). Now, combining these sample sizes from the ISPs’ profiles I think could potentially provide a good basis for estimating overall market share.

My results are presented in Table 1 below. Market share is ALL business, government and home subscribers with ANY kind of internet access including dialup, DSL, cable, fibre, satellite and wireless (fixed & mobile).

The Roy Morgan samples behind these estimates were taken between April 2011 and March 2012 — except where marked with an asterix. Marked samples were taken between April *2010* and March 2012 (i.e. over two years instead of one), so please regard estimates based on these data with extra caution. Market share in these cases could actually be much smaller than calculated here.

Several ISPs listed individually in Table 1 are actually subsidiaries of larger groups, particularly the iiNet group. As a group, iiNet now has a combined market share of 13.2%. Please refer to the notes under the table for more details.

**Table 1: Estimated Australian ISP market share, March 2012**

Internet Service Provider | Roy Morgan sample size (no.) |
Estimated relative market share (%) |
---|---|---|

3 Internet | 243 | 1.8% |

AAPT | 255 | 1.9% |

Adam | 185 | 1.4% |

Chariot | 135 * | 1.0% * |

Dodo | 348 | 2.6% |

Exetel | 119 | 0.9% |

iiNet | 660 | 4.9% |

Internode | 404 | 3.0% |

iPrimus | 237 | 1.8% |

Netspace | 164 | 1.2% |

Optusnet | 2,367 | 17.6% |

TADAust Connect | 144 * | 1.1% * |

Telstra Bigpond | 6,607 | 49.2% |

TPG | 749 | 5.6% |

Unwired | 111 * | 0.8% * |

Virgin | 163 | 1.2% |

Vodafone | 252 | 1.9% |

Westnet | 297 | 2.2% |

TOTAL |
13,440 |
100.0% |

* Treat these data/estimates with extra caution. Time period for these samples are two years, April 2010 – March 2012. All other time periods are one year, April 2011 – March 2012.

**Important notes:**

- AAPT, Netspace, Westnet, and Internode owned by iiNet.
**Estimated market share for whole iiNet group = 13.2%.** - Adam Internet serves South Australia and Northern Territory only. Also, Adam are in the process of being acquired by Telstra (at time of writing).
- Chariot is owned by TPG.

——

]]>

For example, you can see here how frequently the numbers 1 through 45 have been drawn in the Saturday X Lotto. At the time of writing this, **Number 8** was the most frequently drawn, recorded as occurring a total of 289 times between Draw Number 351 to 3265. Note that the Saturday X Lotto Draws are odd-numbered, so Draw 351 to 3265 actually consists of 1458 (i.e. (3265-351)/2+1) weekly games.

Of course there will be random variation in how frequently balls are drawn over time, just as there’s random variation in heads and tails in the toss of a coin. But is it particularly unusual for the Number 8 Ball being drawn 289 times in 1458 games of Saturday X Lotto?

Is South Australia’s Saturday X Lotto biased towards the number 8?

Now before we can determine if the Number 8 being drawn 289 times in 1458 games of X Lotto is an extraordinary event, it helps if we first work out how many times we *expected* it to happen. In X Lotto, a total of eight balls (6 balls for the main prize and then 2 supplementary balls) are selected without replacement from a spinning barrel of 45 balls. The probability of any single number of interest being selected in eight attempts without replacement from a pool of 45 can be calculated using a Hypergeometric Calculator as P(X=1)=0.17778 (i.e. just under 18%). Therefore we expect the Number 8 (or any other number for that matter) to be drawn 0.17778 x 1458 = 259 times in 1458 games.

So observing 289 occurrences when we were only expecting 259 certainly seems unusual, but is it extraordinary?

To answer this, I’ll employ the Binomial test to evaluate the null hypothesis, H0:

H0: Observed frequency of the Number 8 being drawn in SA Lotteries’ Saturday X Lotto is within the expected limits attributable to chance (i.e. the lotto is a fair draw)

vs. the alternative hypothesis, H1:

H1: Observed frequency is higher than would be expected from chance alone (i.e. the lotto is not a fair draw)

The statistics package, R, can be used to run the Binomial test:

> binom.test(289,1458,0.17778,alternative=”greater”)

> Exact binomial test

> data: 289 and 1458

> number of successes = 289, number of trials = 1458, p-value = 0.02353

> alternative hypothesis: true probability of success is greater than 0.17778

So we can reject the null hypothesis of a fair draw at the alpha=0.05 level of significance. The p-value is small enough to **conclude that the true probability of Number 8 being drawn is higher than expected** based on chance alone.

**However**, please note that I am **definitely not** suggesting that anything untoward is going on at SA Lotteries, or that you’ll improve the odds of winning the lottery by including the number 8 in your selection. For a start, rejection of the null hypothesis of a fair system occurs at the standard, but fairly conservative, alpha=0.05 level. What if I had decided to use alpha=0.01 instead? The null hypothesis of a fair system would be retained. Things are all rather arbitrary in the world of statistics.

Still, a curious result that utilised several statistical concepts that I thought would be interesting to blog about.

——

]]>

Amongst other things, it shows deductions claimed by taxpayers for gifts or donations to charities — including welfare agencies, hospitals, research institutes, environmental groups, and arts organisations. From these data I thought it might be interesting to see which of the Australian States and Territories is the most relatively charitable — using tax deductions claimed as a proxy for actual donations made.

I used the Selected items, by sex and State/Territory of residence, 2009-10 income year data to compare taxpayers’ deductions claimed for gifts or donations to charities, relative to their total incomes. The average income per taxable individual in Australia was $66,502 per annum in 2009-10, of which an average $216 (proportionally, 0.32% of income) was claimed for charity. The States and Territories are summarised in Table 1 below.

**Table 1: Charity as a proportion of income, States & Territories, 2009-10**

The data in the table are graphed below. The red horizontal line is the Australian average against which we can compare the individual States and Territories.

**Figure 1: ****Charity as a proportion of income, States & Territories, 2009-10**

New South Welshpersons and Australian Capital Territorians really pull their weight when it comes to donating to charity. They have some of the highest average individual incomes in the country ($69,431 p.a. and $72,007 p.a., respectively), but then really come to the party with the highest proportion of that income going to charitable organisations (0.40% or $277, and 0.39% or $279, respectively).

Western Australians on the other hand really need to lift their game. Despite an average annual individual income of $71,690 (second highest in the country), they were only half as charitable as their eastern cousins mentioned above — only 0.22% ($161) of their income went to charity in the 2009-10 financial year. The Northern Territory recorded the lowest proportion of income to charity with 0.19% of $64,527 ($124) claimed as gifts and donations.

Or perhaps they actually gave a lot to charity, but then didn’t claim it as a deduction come tax-time. Statistics can be a bit dodgy like that.

Interesting.

——

]]>

It’s an interesting read, but the author makes a lot of basic errors. Unfortunately, customers refuse to line up and call at regular intervals and spend the average amount of time on the call. The reality is obviously more bursty than that and needs non-linear modelling.

- Michael Malone, CEO of iiNet

The feedback from Michael Malone above was in response to my previous blog post on *Applying Queuing Theory to iiNet Call Centre Data*. I don’t accept that I made “a lot of basic errors”, but I *did* make a lot of assumptions. Or perhaps the statistician George E. P. Box said it better, “Essentially, all models are wrong, but some are useful.”

But Michael is correct – customers don’t line up and call at regular intervals, and the reality *is* more “bursty” (i.e. Poisson). My model *is* inadequate because it doesn’t take into account all the natural variation in the system.

One way of dealing with, or incorporating, this random variation into the model is by applying Monte Carlo methods.

Take the iiNet Support Phone Call Waiting Statistics for 6 February 2012, specifically for the hour 11am to noon. I chose this time block because the values are relatively easy to read off the graph’s scale – (a bit over) 664 calls and an average time in the queue of 24 minutes.

Now if we assume Average Handling Time (AHT), including time on the call itself followed by off-phone wrap-up time, was 12 minutes, then my model says there were 664*(12/60) / (24/60 +1) = 95 iiNet Customer Service Officers (CSOs) actually taking support calls between 11am and noon on 6 February 2012. That’s an estimate of average number of CSOs *actually on the phones and taking calls during that hour*, excluding those on a break, performing other tasks, and so on. Just those handling calls.

But there will be *a lot* of variation in conditions amongst those 664 calls. I constructed a little Monte Carlo simulation and ran 20,000 iterations of the model with random variation in call arrival rates, AHT, and queue wait times.

Assumptions:

Little’s Law applies

664 calls were received that hour (at a steady pace)

Average time in the queue of 24 minutes

AHT (time on the actual call itself plus off-call wrap-up) of 12 minutes

then the result of the 20,000 monte carlo runs is a new estimate of **135 iiNet CSOs taking support calls between 11am and noon on 6 February 2012**.

I ran a few more simulations, plugging in different values for number of CSOs handling calls (all else remaining equal – i.e. 664 calls an hour; AHT=12 minutes) to see what it did for average time in the queue. The results are summarised in the table below:

Modelling suggests that if iiNet wanted to bring the average time in the phone call support queue down to a sub-5 minute level during that particular hour of interest, an additional 85% in active phone support resourcing would need to be applied.

The table of results is graphically presented below (y-axis is time in queue, x-axis is CSOs)

Looks nice and non-linear to me You can see a law of diminishing returns thing start to take place around about the point of the graph corresponding to 160 CSOs / 16.5 minute average queue wait time.

——

]]>

The long-term average number of support staff (N) in a stable system is equal to the total number of customers in the queue (Q) multiplied by the average time support staff spend resolving a customers’ technical problems (T), divided by the total time waited in the queue (W); or expressed algebraically: N=QT/W.

Thinking things through a bit more, the total number of customers in the queue (Q) at a point in time in a stable system should be equal to the rate at which people joined the queue (* λ*), minus the rate at which the support desk dealt with technical problems (i.e. N/T) over the period of observation. Obviously Q>=0.

So N=QT/W and Q=* λ*-N/T which all comes out in the wash as:

**N=λT/(W+1)**

I thought might be a bit of fun to see if this could be applied to the customer support call centre waiting statistics published by one of Australia’s largest ISPs, *iiNet*.

iiNet make some support centre data available via their customer toolbox page. Below is a screenshot of call activity and wait times graphed each hour by iiNet on 10 January 2012. The green line (in conjunction with the scale on the left hand side of the graph) represents the average time (in minutes) it took to speak to a customer service representative (CSR), including call-backs. The grey bars (in conjunction with the right hand scale) represents the total number of incoming phone calls to iiNet’s support desk.

It may be possible to use the formula derived above to estimate how many CSRs iiNet had on the support desk handling calls that day. For example, during the observed peak period of 8am to 1pm on Tuesday, 10 January 2012, the iiNet support desk was getting around 732 calls per hour on average. The expected wait time in the queue over the same period was around 11 minutes.

If we assume that the average time taken for a CSR to resolve a technical problem is, let’s say, 12.5 minutes, then we can estimate that the number of CSRs answering calls in a typical peak-hour between 8am to 1pm on 10 January 2012 as:

732*(12.5/60) / (11/60 + 1)

= 129 CSRs actively handling calls.

Sounds sort of a reasonable for a customer service-focussed ISP the size of iiNet. But if iiNet wanted to bring the average time in the queue down even more – to a more reasonable 3 minutes, for example – they’d need 145 CSRs (all else remaining equal) during a typical peak-hour answering calls.

————

]]>

I updated my “Stock Market Seismometer” (click on the separate tab above for details) for the first time in many months. I have to say, the results shocked me. Over the course of 2011 the Australian stock market slid down even further into “oversold” territory. As we head into 2012 things have never looked so bleak. Or maybe 2012 will be the year of the rebound? I expect at some point growth will move back to its long term trend, but when that will start to happen is anyone’s guess.

——

]]>

Poker machines are unique in the gambling world. They are the only form of gambling that has been designed and crafted for the purpose of making money, and where there is absolutely no chance of influencing the outcome.

- Tom Cummings, *Poker machine maths,* 27 May 2011

The article, linked above, on poker machines is an excellent insight into the mathematics of poker machine gaming and how, despite a guaranteed return-to-player percentage which is high, a punter that plays long enough ends up with nothing. The vital point that Tom Cummings makes is this:

It’s common knowledge that poker machines have a return-to-player percentage of anywhere between 85 per cent and 90 per cent, depending on where you live and what kind of establishment you’re playing in. For the sake of the story, let’s assume that Gladys’ poker machine was set to 90 per cent. That means that over a long period of time, the machine will return 90 per cent of money gambled to players, and keep 10 per cent as profit.

But wait, I hear you cry. Gladys didn’t lose 10 per cent; she lost it all! Well, no… not according to poker machine mathematics. The rule is that the poker machine has to return 90 per cent of money gambled… not money inserted.

And there’s a huge difference.

The way it works, mathematically, is, I think, very interesting. Take the example used in the article of a 2-cent gaming machine, at $1 per play, with guaranteed return of 90%. When you load that dollar in the slot, sometimes you win, and sometimes you lose, but the long-term average dictates that at each “turn” the dollar is losing 10% of its value. $1 turns into 90c, turns into 81c, turns into 72.9c, … , turns into 2c at which point the game ends.

More generally, at “turn” n, an amount A_{1} initially fed into a slot machine with return R is worth:

or solving for n:

What does this mean? It means that it takes, on average

n = [ *ln*(0.02) - *ln*(1.00) / *ln*(0.90) ] + 1 = 38

iterations to turn $1.00 into 2 cents on a 90% return poker machine. More importantly, the total amount gambled isn’t $1.00, it’s actually (because you’re re-investing your winnings and following your losses until the game ends): $1.00 + 90c + 81c + 72.9c + … + 2c. Or more generally,

which you’ll remember converges to:

So a single dollar coin will generate, on average (on a 2-cent, 90% return machine):

[ 1-0.9^{(38-1)}] / [ 1-0.90 ] = $9.80 in total bets.

At $1.00 per game, at 10 games per minute, that’s about 1 minute of play for every dollar put in, or 5 hours to burn through Gladys’ $300.

Those who defend poker machines often point to the high rate of return as one of the reasons that pokies are just “good, clean fun” for most people. The reality is that every poker machine can meet this “rate of return” requirement while still leaving the gambler broke. That’s poker machine mathematics.

——

Further reading:

ABC Hungry Beast, The Beast File: Pokies

]]>

Australia’s 43rd parliament is a precarious one. Last year’s federal election resulted in a hung parliament in the House of Representatives, with a minority Labor government clinging to power with the help of the Greens and a few Independents. Now I don’t want to sound morbid but, Heaven forbid, should any one of the Honorable Members die, the resulting by-election could conceivably bring down the government. So finely balanced are the numbers.

So, I wonder, what IS the probability that this will happen before the next election, due in 2013?

Statisticians that devote themselves to thinking about exactly this sort of question are called Actuaries, and when calculating risk of mortality they reference a thing called a Life Table. Life tables list the probabilities that a person aged *x* will survive to *x*+1 years. For example, looking at the most recent *Life Tables for Australia*, the probability that a female aged 40 will die before turning 41 is 0.00078 (i.e. 0.078%). Or looking at it in a more positive light, the chance a 40 year old female will live to see her 41st birthday is 1-0.00078, or a healthy 99.922%.

So turning our attention back to the 150-member House of Representatives, what is the probability that one or more of them might shuffle off this mortal coil in the next two years, resulting in a by-election and/or a change of government?

To be rigorous I should aggregate individual probabilities of survival based on each member’s age and gender. But I can’t be arsed, so I’ll talk about it in terms of generalities.

If I recall correctly, the average age of a politician is 51 years. Despite this being a modern society in the year 2011, our parliament is still a massive sausage-fest. Men in government significantly outnumber women. So I’ll just be lazy and use the qx column in the Australian Life Tables linked above for males aged 51 and 52.

From the Life Table, the probability that a male aged 51 years will survive to his 53rd birthday is (1-0.00332)*(1-0.00359)=0.993. Therefore the chance that *every* member of the 150 seat parliament survives for the next two years can be approximated at 0.993^150=0.349 (i.e. 34.9%). The likelihood that one or more parliamentarian dies within the next two years is the complement all surviving, calculated as 1-0.349=0.651 (65.1%).

65%. That’s a pretty high risk that the parliament won’t see out its full term.

——

]]>

The above headline and corresponding story appeared in *ARN* on 4 May 2011. The article was reporting on the Telecommunications Industry Ombudsman’s (TIO) January to March 2011 statistics published here.

Now before I go any further, let me say I have grave concerns with the source data itself, let alone the way it’s being reported in the press. When talking about TIO complaint statistics, it is worthwhile to familiarise yourself with this recent research report from the University of Technology, Sydney (UTS).

**Findings of the UTS research include**** (from the executive summary):**

**Over 80%**(of carriage service providers surveyed) assert that the TIO**accepts complaints which are out of jurisdiction, frivolous or vexatious**.

**Recommendations include:**

- Amend TIO policy and procedure to
**cease the multiple counting of complaints in statistics**and recommence reporting disposition of complaints. - Amend TIO policy and procedure to
**refer to Level 1 Complaints as Contacts rather than Complaints**.

So fair to say the TIO isn’t exactly highly regarded by the ISP industry. Indeed, Exetel have commenced legal action against the TIO for acting outside its charter. Or, in Exetel CEO John Linton’s own words, *“Exetel is now going to take a court action to have the TIO closed as being a ‘criminal’ organisation based on a level of incompetence, lack of knowledge and unconstitutional actions that even Australians, who are mostly as apathetic as the governments they tolerate, shouldn’t have imposed on them.”*

Well OK then…

Now if you still want to accept that all “complaints”, as reported by the TIO, are real, legitimate complaints then the fairest way to present the numbers is as a proportion of overall customer base. That’s going to be a bit tricky, because not all ISPs publish accurate customer numbers in the public domain. So I’ve had to estimate relative market share using Roy Morgan survey data and a methodology used in a previous blog post. When I refer to market share I mean ALL business, government AND home subscribers with ANY kind of internet access including dialup, DSL, cable, fibre, satellite and wireless (fixed & mobile). The Roy Morgan surveys were conducted between January 2010 and December 2010

Then, to be consistent, I only looked at TIO complaints made against a provider’s internet services . That is, I excluded complaints made against landline and mobile phone services. I considered all internet services TIO complaints made between October 2010 and March 2011

Then I derived a normalised score by dividing the relative market share by the proportion of complaints. The benchmark is a score of 1000 and a higher score is better. A score *over 1000* means that an ISP recorded relatively *fewer complaints* as a proportion of its customer base than its peers. Less than 1000 means that an ISP was over-represented in TIO complaints relative to its size.

The TIO complaints leaderboard comes out as follows (click to up-size):

**Table 1: TIO internet services complaints leaderboard, 2010/2011**

iiNet did indeed do very well. With roughly an 11% market share, but only 6% of complaints, it gets a score of 1843. That’s well above the expected benchmark of 1000. But it didn’t fare best. The ISP that recorded the *lowest number of TIO complaints as a proportion of number of customers was Internode*, with a massive normalised score of 3448.

——

]]>

CVC is bandwidth that ISPs will need to purchase from NBN Co, charged at the rate of $20 (ex-GST) per Mbps per month. Note that this CVC is* on top of* the backhaul and international transit required to pipe all those interwebs into your home. And just like backhaul and international transit, if an ISP doesn’t buy enough CVC from NBN Co to cover peak utilisation, its customers will experience a congested service.

The problem with the CVC tax, priced as it is by NBN Co, is that it punishes small players. By my calculations, an ISP of (say) 1000 subscribers will need to spend *proportionally* a lot more on CVC than an ISP of 1,000,000 subscribers if they want to provide a service that delivers the speeds it promises.

Here comes the statistics.

Consider NBN Co’s theoretical 12 megabit service with 10GB of monthly quota example that they use in the document I linked to above. 10GB per month, at 12Mbps gives you 6,827 seconds (a bit under 2 hours) at full speed before you’re throttled off. There’s 2,592,000 seconds in a 30-day month, so if I choose a random moment in time there is a 6827/2592000 = 0.263% chance that I’ll find you downloading at full speed.

That’s on average. The probability would naturally be higher during peak times. But let’s assume in this example that our theoretical ISP has a perfectly balanced network profile (no peak or off-peak periods). It doesn’t affect the point I’ll labour to make.

A mid-sized ISP with (let’s say) 100,000 subscribers can expect, on average, to have 100,000*0.263% = only 263 of those customers downloading at full speed simultaneously at any particular second. However, the Binomial distribution tells us that there’s a relatively small, but still statistically significant (at the alpha=0.05 level) probability that there could be *290 or more* customers downloading at the same time.

So a quality ISP of 100,000 subscribers will plan to buy enough CVC bandwidth to service 263 customers at any one time. But a statistician would advise the ISP to buy enough CVC bandwidth to service *290* subscribers, an *additional* (290-263)/263 = 10% , or find itself with a congested service about one day in every 20.

This additional “burst headroom”, as a percentage, *increases* as the size of the ISP *decreases*. From above, an ISP of 10,000 can expect to have 26 customers downloading simultaneously at any random moment in time. But there’s a statistically significant chance this could be 35+. This requires them to buy an additional (35-26)/26=**33%** in CVC over and above what was expected to cover peak bursts.

The table below summarises, for ISPs of various sizes, how much *additional* CVC would need to be purchased* over and above the expected amount*, to provide an uncontended service 95%+ of the time.

Graphically it looks a bit like this…

As you can see, things only really start to settle down for ISPs larger than 100,000 subscribers. Any smaller than that and your relative cost of CVC per subscriber per month is disproportionally large.

….

Further reading:

Rebalancing NBNCo Pricing Model

NBN Pricing – Background & Examples, Part 1

——

]]>